CBSE 2023 · Region 1 · Set 1 · Q32 · 5 marks
A relation $\displaystyle \mathrm{R}$ is defined on a set of real numbers $\displaystyle \mathbb{\mathrm{R}}$ as \[\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x} \cdot \mathrm{y} \text { is an irrational number }\} . \] Check whether $\displaystyle \mathrm{R}$ is reflexive, symmetric and transitive or not.
Marking-scheme solution
For reflexive
$\displaystyle (1,1) \notin \mathrm{R}$ as $\displaystyle 1^{2}$ is rational
R is not reflexive
For symmetric
Let $\displaystyle (\mathrm{x}, \mathrm{y}) \in \mathrm{R} \therefore \mathrm{x} . \mathrm{y}$ is an irrational number
∴ ( $\displaystyle \mathrm{y} . \mathrm{x}$ ) is an irrational number
$\displaystyle \therefore(\mathrm{y}, \mathrm{x}) \in \mathrm{R}$
$\displaystyle \therefore \mathrm{R}$ is symmetric
For Transitive
$\displaystyle (1, \sqrt{2}) \in \mathrm{R},(\sqrt{2}, 2) \in \mathrm{R} \quad$ but $\displaystyle (1,2) \notin \mathrm{R}$
$\displaystyle \therefore \mathrm{R}$ is not transitive
Relations and FunctionsTypes of RelationsAnalyselong_answermedium
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